Find the area of the region enclosed by the three graphs y = x^2, y = 8 x, and y = 1

To find the area of the region enclosed by the three graphs \( y = x^2 \), \( y = 8x \), and \( y = 1 \), we need to determine the points of intersection between these curves, and then integrate appropriately.

### Step 1: Find the points of intersection

#### Intersection of \( y = x^2 \) and \( y = 8x \):
Set the equations equal to each other:
\[
x^2 = 8x
\]
Rearrange and factor:
\[
x^2 - 8x = 0 \implies x(x - 8) = 0
\]
Thus, \( x = 0 \) or \( x = 8 \).

#### Intersection of \( y = x^2 \) and \( y = 1 \):
Set the equations equal to each other:
\[
x^2 = 1
\]
Taking the square root of both sides:
\[
x = \pm 1
\]

#### Intersection of \( y = 8x \) and \( y = 1 \):
Set the equations equal to each other:
\[
8x = 1
\]
Solve for \( x \):
\[
x = \frac{1}{8}
\]

### Step 2: Determine the relevant intervals and setup integrals

The area of the region is found by integrating the difference between the top curve and the bottom curve over the intervals determined by the points of intersection.

1. **Between \( x = -1 \) and \( x = \frac{1}{8} \):**
   Here, the top curve is \( y = 1 \) and the bottom curve is \( y = x^2 \).
   \[
   A_1 = \int_{-1}^{\frac{1}{8}} (1 - x^2) \, dx
   \]

2. **Between \( x = \frac{1}{8} \) and \( x = 8 \):**
   Here, the top curve is \( y = 8x \) and the bottom curve is \( y = x^2 \).
   \[
   A_2 = \int_{\frac{1}{8}}^{8} (8x - x^2) \, dx
   \]

### Step 3: Evaluate the integrals

#### Evaluate \( A_1 \):
\[
A_1 = \int_{-1}^{\frac{1}{8}} (1 - x^2) \, dx = \left[x - \frac{x^3}{3}\right]_{-1}^{\frac{1}{8}}
\]
Substituting the limits:
\[
A_1 = \left(\frac{1}{8} - \frac{(\frac{1}{8})^3}{3}\right) - \left(-1 - \frac{(-1)^3}{3}\right)
\]
\[
A_1 = \left(\frac{1}{8} - \frac{1}{3 \times 8^3}\right) - \left(-1 + \frac{1}{3}\right)
\]
\[
A_1 = \left(\frac{1}{8} - \frac{1}{1536}\right) - \left(-\frac{2}{3}\right)
\]
\[
A_1 = \left(\frac{1536 - 1}{12288}\right) + \frac{2}{3}
\]
\[
A_1 = \frac{1535}{12288} + \frac{2}{3}
\]

#### Evaluate \( A_2 \):
\[
A_2 = \int_{\frac{1}{8}}^{8} (8x - x^2) \, dx = \left[4x^2 - \frac{x^3}{3}\right]_{\frac{1}{8}}^{8}
\]
Substituting the limits:
\[
A_2 = \left(4 \times 64 - \frac{512}{3}\right) - \left(4 \times \frac{1}{64} - \frac{\left(\frac{1}{8}\right)^3}{3}\right)
\]
\[
A_2 = \left(256 - \frac{512}{3}\right) - \left(\frac{4}{64} - \frac{1}{3 \times 512}\right)
\]
\[
A_2 = 256 - \frac{512}{3} - \left(\frac{1}{16} - \frac{1}{1536}\right)
\]
\[
A_2 = 256 - 170.67 - \left(\frac{96}{1536} - \frac{1}{1536}\right)
\]
\[
A_2 = 256 - 170.67 - \frac{95}{1536}
\]

Finally, the total area is the sum of \( A_1 \) and \( A_2 \).

**Note:** To simplify the calculation, you can substitute the values into the above formulas and add the results.

### Do you need detailed steps to simplify the integrals? Or have any questions? Here are 5 related questions:

1. How do you find the intersection points between two curves?
2. What is the geometric interpretation of the definite integral?
3. How can we determine the top and bottom functions for a region bounded by two curves?
4. What changes if the functions are not continuous over the interval?
5. How would you calculate the area if the region is bounded by more than three curves?

**Tip:** Always sketch the graphs first to better visualize the regions you are integrating.

Learn how to find the area enclosed by the curves y = x^2, y = 8x, and y = 1 using integration. This problem involves determining points of intersection, setting up and evaluating integrals over specific intervals. Suitable for advanced high school students and calculus enthusiasts.

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